Another way to write the Cantor set is to note that each of the sets
A n can be written as a finite union of
2 n closed intervals, each of which has a length of
1 / 3 n, as follows:

A 0 = [0, 1]

A 1 = [0, 1/3]
[2/3, 1]

A 2 = [0, 1/9]
[2/9, 3/9]
[6/9, 7/9]
[8/9, 1]

...

Now suppose that there is an open set U contained in C.
Then there must be an open interval (a, b) contained in C.
Now pick an integer N such that

1 / 3 N &lt b - a

Then the interval (a, b) can not be contained in the
set AN, because that set is comprised of
intervals of length 1 / 3N. But if that interval
is not contained in AN it can not be contained
in C. Hence, no open set can be contained in the Cantor set
C.